NCERT Solutions For Class 11 Maths Chapter 2 Relations And Functions Exercise 2.1 - 2025-26

Exercise 2.1

1. If $\left( \dfrac{x}{3}+1,y-\dfrac{2}{3} \right)=\left( \dfrac{5}{3},\dfrac{1}{3} \right)$ , find the values of  $\text{x}$ and $y$ .

Ans: We are provided with the fact that $\left( \dfrac{x}{3}+1,y-\dfrac{2}{3} \right)=\left( \dfrac{5}{3},\dfrac{1}{3} \right)$ 

These are ordered pairs which are equal to each other, then the corresponding elements should also be equal to each other.

Thus, we will have, $\dfrac{x}{3}+1=\dfrac{5}{3}$ 

And also $y-\dfrac{2}{3}=\dfrac{1}{3}$ 

Now, we will try to simplify the given equations and find our needed values.

$\dfrac{x}{3}+1=\dfrac{5}{3} $ 

 $\Rightarrow \dfrac{x}{3}=\dfrac{5}{3}-1$ 

Simplifying further,

$\Rightarrow \dfrac{x}{3}=\dfrac{5-3}{3}=\dfrac{2}{3}$ 

$\Rightarrow x=2$

So, we have the value of $x$ as $2$.

Again, for the second equation,

 $y-\dfrac{2}{3}=\dfrac{1}{3}$ 

$\Rightarrow y=\dfrac{2}{3}+\dfrac{1}{3}$ 

And, after more simplification,

$\Rightarrow y=\dfrac{1+2}{3}=\dfrac{3}{3}$

$\Rightarrow y=1$

So, we have the values of $x$ and $y$ as $2$ and $1$ respectively.

2. If the set A has 3 elements and the set B={3,4,5}, then find the number of elements in $(A\times B)$.

Ans: We are provided with the fact that the set $A$ has $3$ elements and the set $B$ is given as $\{3,4,5\}$.

So, the number of elements in set $B$ is $3$.

Thus, the number of elements in $(A\times B)$ will be,

= Number of elements in $A\,\times $ Number of elements in $B$ 

$=3\times 3=9$ 

So, the number of elements in $(A\times B)$ is $9$.

3. If $G=\{7,8\}$ and $H=\{5,4,2\}$ , find $G\times H$ and $H\times G$ .

Ans: We have the sets $G=\{7,8\}$ and $H=\{5,4,2\}$ .

The Cartesian product of two non-empty sets $A$ and $B$ is defined as $A\times B=\{(a,b):a\in A\,and\,\,b\in B\}$ 

So, the value of $G\times H$ will be, 

$G\times H=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}$ 

And similarly the value of $H\times G$ will be,

$H\times G=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$ 

4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If $P=\{m,n\}$ and $Q=\{n,m\}$ , then $P\times Q=\{(m,n),(n,m)\}$ .

Ans: The statement is False.

We have the value as, $P=\{m,n\}$ and $Q=\{n,m\}$.

Thus, $P\times Q=\{(m,m),(m,n),(n,m),(n,n)\}$

(ii) If $A$ and $B$ are non-empty sets, then $A\times B$ is a non-empty set of ordered pairs $(x,y)$ such that $x\in A$ and $y\in B$.

Ans: The statement is True.

(iii) If $A=\{1,2\},B=\{3,4\}$ , then $A\times \{B\cap \varnothing \}=\varnothing $ .

Ans: The statement is True.

We know, $B\cap \varnothing =\varnothing $ 

Thus, we have, $A\times \{B\cap \varnothing \}=A\times \varnothing    =\varnothing $.

5. If $A=\{-1,1\}$ , find $A\times A\times A$ .

Ans: For any non-empty set $A$ , the set $A\times A\times A$ is defined by, 

$A\times A\times A=\{(p,q,r):p,q,r\in A\}$  

Now, we are provided with the fact that, $A=\{-1,1\}$ 

Thus, 

$A\times A\times A=\left\{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right\}$

6. If $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$ . Find $A$ and $B$ .

Ans: We are provided with the fact that $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$ 

On the other hand, the Cartesian product of two non-empty sets $A$ and $B$ is defined as $A\times B=\{(a,b):a\in A\, and\,\,b\in B\}$ 

As we can see, $A$ is the set of all the first elements and $B$ is the set of all the second elements.

So, we will have, $A=\{a,b\}$ and $B=\{x,y\}$ .

7. Let $A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\}$ and $D=\{5,6,7,8\}$ . Verify that

(i) $A\times (B\cap C)=(A\times B)\cap (A\times C)$ 

Ans: We are provided with 3 sets and we have to prove $A\times (B\cap C)=(A\times B)\cap (A\times C)$ 

To start with, we will have, $B\cap C=\varnothing $, as there are no elements in common between these sets.

Thus, we have, $A\times (B\cap C)=A\times \varnothing =\varnothing $ 

For the right-hand side, we have,

$A\times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$ 

And similarly,

$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$ 

Again, we can see there are no elements in common between these sets. So, we have, $(A\times B)\cap (A\times C)=\varnothing $ 

So, we get, L.H.S = R.H.S.

(ii) $A\times C$ is a subset of $B\times D$ 

Ans: Again, we are to verify, $A\times C$ is a subset of $B\times D$ 

So, we have, $A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$ 

And similarly,

$B\times D=\{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\}$ 

We can easily see, every element of $A\times C$ is an element of $B\times D$. So, $A\times C$ is a subset of $B\times D$.

8. Let $A=\{1,2\}$ and $B=\{3,4\}$ . Write $A\times B$ . How many subsets will $A\times B$ have? List them.

Ans: We are provided with the fact that $A=\{1,2\}$ and $B=\{3,4\}$ 

Thus, we have, $A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$ 

So, the set $A\times B$ has 4 elements.

Now, this is also known to us that, if a set $A$ has $n$ elements, then the number of subsets of $A$is ${{2}^{n}}$.

We can thus conclude that, $A\times B$will have ${{2}^{4}}=16$ subsets.

Now, noting down the subsets of $A\times B$ we get,

$\varnothing ,\{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\{(1,3),(1,4)\}, $

$\{(1,3),(2,3)\},\{(1,3),(2,4)\}, $

$\{(1,4),(2,3)\},\{(1,4),(2,4)\},\{(2,3),(2,4)\}, $ 

$\{(1,3),(1,4),(2,3)\},\{(1,3),(1,4),(2,4)\}, $ 

$\{(1,3),(2,3),(2,4)\},\{(1,4),(2,3),(2,4)\}, $ 

$\{(1,3),(1,4),(2,3),(2,4)\} $ 

9. Let $A$ and $B$ be two sets such that $n(A)=3$ and $n(B)=2$ . If $(x,1),(y,2),(z,1)$ are in $A\times B$ , find $A$ and $B$ , where $x,y$ and $z$ are distinct elements.

Ans: We are provided with the fact that $n(A)=3$ and $n(B)=2$ ; and\[(x,1),(y,2),(z,1)\] are in $A\times B$ .

We also know that $A$ is the set of all the first elements and $B$ is the set of all the second elements.

So, we can conclude, $A$ having elements $x,y,z$ and $B$ having elements $1,2$ .

Thus, we get, $n(A)=3,n(B)=2$.

So, $A=\{x,y,z\},B=\{1,2\}$ .

10. The Cartesian product $A\times A$ has 9 elements among which are found $(-1,0)$ and $(0,1)$ . Find the set $A$ and the remaining elements of $A\times A$.

Ans: We are provided with, $n(A\times A)=9$.

We also know that, if $n(A)=a,n(B)=b$ , then $n(A\times B)=ab$ 

As it is given that, $n(A\times A)=9$

It can be written as,

 $n(A)\times n(A)=9$ 

 $\Rightarrow n(A)=3$ 

And it is also given that $(-1,0),(0,1)$ are the two elements of $A\times A$ .

Again, the fact is also known that, $A\times A=\{(a, a):a\in A\}$. And also $-1,0,1$ are the elements of $A$.

Also, $n(A)=3$ , implies $A=\{-1,0,1\}$ .

So, $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)$ are the remaining elements of $A\times A$ .

Conclusion

Completing Exercise 2.1 of Chapter 2 in Class 11 Maths provides a solid understanding of the Cartesian Products of Sets. This exercise is essential as it lays the groundwork for more advanced topics in relations and functions. By practising the problems in class 11 ex 2.1, students learn how to form and interpret ordered pairs from two sets, which is a fundamental skill in set theory. Understanding these concepts prepares students for further studies in mathematics, enhancing their ability to analyse and work with complex relationships between sets.

Class 11 Maths Chapter 2: Exercises Breakdown

CBSE Class 11 Maths Chapter 2 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for CBSE Class 11 Maths